Physics forces problem.?
While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 34.5 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance required is about 1.9 km.
(a) If the mass of a jet is 1.57 multiplied by 105 kg, what force is needed for takeoff?
Comments
I assume you mean the mass is 1.57 * 10^5 kg, which makes a BIG difference.
Find the acceleration:
s = ut + 0.5at^2
1900 = 0 + 0.5*a*34.5^2
1900 = 595.125a
a = 1900 / 595.125
a = 3.19 m/s^2
F = ma
F = 1.57 * 10^5 * 3.19
F = 5.01 * 10^5 N to three significant figures.
Edit: Amaya, you make so many mistakes there I'm not sure where to begin.
Sum Fx = zero and Sum Fy = zero Sum Fx = F1,x + F2,x + F3,x = zero + forty four cos 60 + F3,x = 22 + F3,x = zero. So, F3,x = -22 N. Sum Fy = F1,y + F2,y + F3,y = 33 + forty four sin 60 + F3,y = 33 + forty four sin 60 + F3,y = zero So, F3,y = -seventy one N. F3 = sqrt [ F3,x ^two + F3,y ^two ] = sqrt [ (-22)^two + (-seventy one)^two ] = seventy four N. Direction of F3 = a hundred and eighty + tan^(-one million) (seventy one/22) = a hundred and eighty + seventy three = 253 deg.
u=0
v=1.9/34.5 km/s
t=34.5 s
a= (v-u)/t = 0.0016m/s^2
f=ma = 1.57 x 105 x 0.0016
ans = 0.26376 N